And because the straight line is tangent to the circle x 2+y 2 = 4,

So the distance from the center of the circle to the straight line is exactly radius 2.

So | 4 |/√ (1+k 2) = 2.

The solution is k = √ 3,

Substitute the linear equation y = √ 3x+4 into y 2 = 2px.

3x^2 8√3x+ 16=2Px

Let A(x 1, y 1)B(x2, y2).

Then x 1x2= 16/3,

Because y12 = 2px1and y2^2=2Px2.

So (y1y2) 2 = 4p2 * x1x2 = 64p2/3.

So y 1y2 = 8 √ 3p/3.

Because OA is vertical OB

So x 1x2+y 1y2=0.

So 16/3 8 √ 3p/3 = 0.

So p = 2 √ 3/3.

Because of P>0, P=2√3/3.

So the equation of parabola is y 2 = 4 √ 3x/3.

2. Because the center of the ellipse is at the origin O and the focus is on the X axis, let the ellipse be x 2/a 2+y 2/b 2 =1.

So a>b>0

Because the eccentricity is e = √ 3/2 = c/a = √ (a 2-b 2)/a 2-b 2)/a.

So a=2b

Because the point (√2, √2/2) is on the ellipse, it can be substituted into the elliptic equation to solve it.

B= 1, so a=2.

So the elliptic equation is x 2/4+y 2 =1.

Suppose that the straight line L with the origin o intersects the ellipse at two points: P(x 1, y 1) and q (x2, y2).

Let the straight line l be y=kx+b and substitute it into the elliptic equation.

simplify

( 1+4k^2)x^2+8bkx+4(b^2- 1)=0

So x1x 2 = 4 (B2-1)/(1+4k2).

x 1+x2=-8bk/( 1+4k^2)

Therefore, y1+y2 = kx1+b+kx2+b = k (x1+x2)+2b = 2b/(1+4k2).

y 1y2=(kx 1+b)(kx2+b)=k^2x 1x2+bk(x 1+x2)+b^2=(b^2-4k^2)/( 1+4k^2)

Because the slope of the straight line OP is y 1/x 1, the slope of the straight line PQ is k=(y2-y 1)/(x2-x 1).

The slope of the straight line OQ is y2/x2.

And the slOPes of the straight lines op and PQ. OQ increases geometrically in turn.

So (y1/x1) (y2/x2) = k 2.

So y 1y2 = k 2x 1x2.

So (B2-4k2)/(1+4k2) = 4k2 (B2-1)/(1+4k2).

So k 2 = 1/4.

So the straight line l: y = x/2+b

Let d be the distance from point o to line PQ.

So d = | b |/√ (1+k 2).

Because | x2-x1| = √ [x1+x2] 2-4x1x2] = √ [x1+x2) 2-4x1x2] = √ (8-x2)

|pq|=|x2-x 1|*√( 1+k^2)=√(8-4b^2)*√( 1+k^2)

So s △ opq =1/2 * d * | pq | =1/2 * | b | * √ (8-4b 2).

Let m = | b | * √ (8-4b 2)

So m 2 = b 2 * (8-4b 2)

Simplified to 4b 4-8b 2+m 2 = 0.

From δ = 64-4 * 4 * m 2 ≥ 0

M≤2。

Because m is non-negative, M≥0.

If M=0, then b=0 or b 2 = 2.

When b=0 is yes, the straight line l passes through the origin, which does not conform to the meaning of the question.

When b 2 = 2, (1+4k2) x2+8bkx+4 (B2-1) = 0 can be changed to (x √ 2) 2 = 0.

At this time, p and q are the same point, which does not conform to the △OPQ in the meaning of the question.

So 0

So 0